Integrand size = 96, antiderivative size = 29 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=x+(2-x) \left (5+\frac {2}{\left (3-\frac {x}{5}\right ) \log (\log (\log (2 x)))}\right ) \] Output:
(2-x)*(5+2/(3-1/5*x)/ln(ln(ln(2*x))))+x
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-2 \left (2 x-\frac {5 (-2+x)}{(-15+x) \log (\log (\log (2 x)))}\right ) \] Input:
Integrate[(-300 + 170*x - 10*x^2 - 130*x*Log[2*x]*Log[Log[2*x]]*Log[Log[Lo g[2*x]]] + (-900*x + 120*x^2 - 4*x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2 *x]]]^2)/((225*x - 30*x^2 + x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]] ^2),x]
Output:
-2*(2*x - (5*(-2 + x))/((-15 + x)*Log[Log[Log[2*x]]]))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {-10 x^2+\left (-4 x^3+120 x^2-900 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+170 x-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))-300}{\left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-10 x^2+\left (-4 x^3+120 x^2-900 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+170 x-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))-300}{x \left (x^2-30 x+225\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int -\frac {5 x^2+65 \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x))) x-85 x+2 \left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+150}{2 (15-x)^2 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -2 \int \frac {5 x^2+65 \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x))) x-85 x+2 \left (x^3-30 x^2+225 x\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))+150}{(15-x)^2 x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -2 \int \left (\frac {5 (x-2)}{(x-15) x \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}+\frac {65}{(x-15)^2 \log (\log (\log (2 x)))}+2\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \left (\frac {13}{3} \int \frac {1}{(x-15) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}dx+65 \int \frac {1}{(x-15)^2 \log (\log (\log (2 x)))}dx+2 x-\frac {2}{3 \log (\log (\log (2 x)))}\right )\) |
Input:
Int[(-300 + 170*x - 10*x^2 - 130*x*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x] ]] + (-900*x + 120*x^2 - 4*x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]]^ 2)/((225*x - 30*x^2 + x^3)*Log[2*x]*Log[Log[2*x]]*Log[Log[Log[2*x]]]^2),x]
Output:
$Aborted
Time = 4.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(-4 x +\frac {-20+10 x}{\left (x -15\right ) \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}\) | \(23\) |
| parallelrisch | \(\frac {-300-60 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) x^{2}+150 x +13500 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right )}{15 \ln \left (\ln \left (\ln \left (2 x \right )\right )\right ) \left (x -15\right )}\) | \(40\) |
Input:
int(((-4*x^3+120*x^2-900*x)*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))^2-130*x*ln (2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))-10*x^2+170*x-300)/(x^3-30*x^2+225*x)/ln( 2*x)/ln(ln(2*x))/ln(ln(ln(2*x)))^2,x,method=_RETURNVERBOSE)
Output:
-4*x+10*(-2+x)/(x-15)/ln(ln(ln(2*x)))
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \] Input:
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm="f ricas")
Output:
-2*(2*(x^2 - 15*x)*log(log(log(2*x))) - 5*x + 10)/((x - 15)*log(log(log(2* x))))
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=- 4 x + \frac {10 x - 20}{\left (x - 15\right ) \log {\left (\log {\left (\log {\left (2 x \right )} \right )} \right )}} \] Input:
integrate(((-4*x**3+120*x**2-900*x)*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))**2 -130*x*ln(2*x)*ln(ln(2*x))*ln(ln(ln(2*x)))-10*x**2+170*x-300)/(x**3-30*x** 2+225*x)/ln(2*x)/ln(ln(2*x))/ln(ln(ln(2*x)))**2,x)
Output:
-4*x + (10*x - 20)/((x - 15)*log(log(log(2*x))))
Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-\frac {2 \, {\left (2 \, {\left (x^{2} - 15 \, x\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right ) - 5 \, x + 10\right )}}{{\left (x - 15\right )} \log \left (\log \left (\log \left (2\right ) + \log \left (x\right )\right )\right )} \] Input:
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm="m axima")
Output:
-2*(2*(x^2 - 15*x)*log(log(log(2) + log(x))) - 5*x + 10)/((x - 15)*log(log (log(2) + log(x))))
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=-4 \, x + \frac {10 \, {\left (x - 2\right )}}{x \log \left (\log \left (\log \left (2 \, x\right )\right )\right ) - 15 \, \log \left (\log \left (\log \left (2 \, x\right )\right )\right )} \] Input:
integrate(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)) )^2-130*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3 -30*x^2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x, algorithm="g iac")
Output:
-4*x + 10*(x - 2)/(x*log(log(log(2*x))) - 15*log(log(log(2*x))))
Time = 3.92 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=\frac {2\,\left (5\,x-2\,x^2\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )+30\,x\,\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )-10\right )}{\ln \left (\ln \left (\ln \left (2\,x\right )\right )\right )\,\left (x-15\right )} \] Input:
int(-(10*x^2 - 170*x + log(log(log(2*x)))^2*log(2*x)*log(log(2*x))*(900*x - 120*x^2 + 4*x^3) + 130*x*log(log(log(2*x)))*log(2*x)*log(log(2*x)) + 300 )/(log(log(log(2*x)))^2*log(2*x)*log(log(2*x))*(225*x - 30*x^2 + x^3)),x)
Output:
(2*(5*x - 2*x^2*log(log(log(2*x))) + 30*x*log(log(log(2*x))) - 10))/(log(l og(log(2*x)))*(x - 15))
Time = 0.17 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {-300+170 x-10 x^2-130 x \log (2 x) \log (\log (2 x)) \log (\log (\log (2 x)))+\left (-900 x+120 x^2-4 x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))}{\left (225 x-30 x^2+x^3\right ) \log (2 x) \log (\log (2 x)) \log ^2(\log (\log (2 x)))} \, dx=\frac {-4 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2 x \right )\right )\right ) x^{2}+60 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2 x \right )\right )\right ) x +10 x -20}{\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2 x \right )\right )\right ) \left (x -15\right )} \] Input:
int(((-4*x^3+120*x^2-900*x)*log(2*x)*log(log(2*x))*log(log(log(2*x)))^2-13 0*x*log(2*x)*log(log(2*x))*log(log(log(2*x)))-10*x^2+170*x-300)/(x^3-30*x^ 2+225*x)/log(2*x)/log(log(2*x))/log(log(log(2*x)))^2,x)
Output:
(2*( - 2*log(log(log(2*x)))*x**2 + 30*log(log(log(2*x)))*x + 5*x - 10))/(l og(log(log(2*x)))*(x - 15))